Analytic functions and diagonalisation of matrices.












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$begingroup$


If I have an analytic function $f$ of a square matrix A (like sin(A)), then I know that if the matrix diagnosable then it is possible to find a matrix
$$D = P^{-1}AP tag{1}$$. Then for a function $f(A)$:
$$f(A) = Pf(D)P^{-1} tag{2}$$



So for example if $f(x) = cos(x)$ :
$$cos(A) = Pf(D)P^{-1} tag{3}$$



What is the justification for this and how does this follow from first principles? A similar question has been asked here $sin(A)$, where $A$ is a matrix but I don't see any justification/proof for (2).






linear-algebra matrices matrix-calculus diagonalization







share|cite|improve this question











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    0












    $begingroup$


    If I have an analytic function $f$ of a square matrix A (like sin(A)), then I know that if the matrix diagnosable then it is possible to find a matrix
    $$D = P^{-1}AP tag{1}$$. Then for a function $f(A)$:
    $$f(A) = Pf(D)P^{-1} tag{2}$$



    So for example if $f(x) = cos(x)$ :
    $$cos(A) = Pf(D)P^{-1} tag{3}$$



    What is the justification for this and how does this follow from first principles? A similar question has been asked here $sin(A)$, where $A$ is a matrix but I don't see any justification/proof for (2).






    linear-algebra matrices matrix-calculus diagonalization







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      If I have an analytic function $f$ of a square matrix A (like sin(A)), then I know that if the matrix diagnosable then it is possible to find a matrix
      $$D = P^{-1}AP tag{1}$$. Then for a function $f(A)$:
      $$f(A) = Pf(D)P^{-1} tag{2}$$



      So for example if $f(x) = cos(x)$ :
      $$cos(A) = Pf(D)P^{-1} tag{3}$$



      What is the justification for this and how does this follow from first principles? A similar question has been asked here $sin(A)$, where $A$ is a matrix but I don't see any justification/proof for (2).






      linear-algebra matrices matrix-calculus diagonalization







      share|cite|improve this question











      $endgroup$




      If I have an analytic function $f$ of a square matrix A (like sin(A)), then I know that if the matrix diagnosable then it is possible to find a matrix
      $$D = P^{-1}AP tag{1}$$. Then for a function $f(A)$:
      $$f(A) = Pf(D)P^{-1} tag{2}$$



      So for example if $f(x) = cos(x)$ :
      $$cos(A) = Pf(D)P^{-1} tag{3}$$



      What is the justification for this and how does this follow from first principles? A similar question has been asked here $sin(A)$, where $A$ is a matrix but I don't see any justification/proof for (2).






      linear-algebra matrices matrix-calculus diagonalization




      linear-algebra matrices matrix-calculus diagonalization






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 12 at 16:38







      daljit97

















      asked Jan 12 at 16:18









      daljit97daljit97

      178111




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          1 Answer
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          $begingroup$

          If $f(A)$ is defined for every $A$ then $f$ must be an entire function: $$f(z)=sum_{n=0}^infty c_n z^n.$$Note that $$(PDP^{-1})^n=PD^nP^{-1}.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            (+1) Please let me note that the term $f(D)$ or $cos(D)$ is not even defined. You have to first express $f$ or $cos$ by a limit of polynomials in order to give it a meaning.
            $endgroup$
            – Yanko
            Jan 12 at 17:00








          • 1




            $begingroup$
            @Yanko Huh??? A power series is a limit of polynomials. (No, that's not the only way, but it's the simplest, and it certainly applies here.)
            $endgroup$
            – David C. Ullrich
            Jan 12 at 17:03










          • $begingroup$
            Ok so $f(A) = f(PDP^{-1}) = sum_{n=0}^infty c_n (PDP^{-1})^n=Psum_{n=0}^infty c_n (D)^n P^{-1}$?
            $endgroup$
            – daljit97
            Jan 12 at 17:27












          • $begingroup$
            @daljit97 $dots=Pf(D)P^{-1}$. Right.
            $endgroup$
            – David C. Ullrich
            Jan 12 at 17:32






          • 1




            $begingroup$
            @Yanko You need to read about the "functional calculus" somewhere - it's a long story. There's the "continuous functional calculus": If $D$ is diagonal with entriies $lambda_1,dots,lambda_n$ then $f(D)$ is diagonal with entries $f(lambda_1),dots,f(lambda_n)$. Or the holomorphic functional calculus: If $x$ is an element of any Banach algebra, $V$ is open, $sigma(x)subset V$ and $fin H(V)$ then you can define $f(x)$ using the Cauchy Integral Formula $f(x)=frac1{2pi i}int_Gamma f(z)(x-ze)^{-1},dz$ for a suitable cycle $Gamma$ (see Rudin Functional Analysis).
            $endgroup$
            – David C. Ullrich
            Jan 12 at 17:51













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          1 Answer
          1






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          $begingroup$

          If $f(A)$ is defined for every $A$ then $f$ must be an entire function: $$f(z)=sum_{n=0}^infty c_n z^n.$$Note that $$(PDP^{-1})^n=PD^nP^{-1}.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            (+1) Please let me note that the term $f(D)$ or $cos(D)$ is not even defined. You have to first express $f$ or $cos$ by a limit of polynomials in order to give it a meaning.
            $endgroup$
            – Yanko
            Jan 12 at 17:00








          • 1




            $begingroup$
            @Yanko Huh??? A power series is a limit of polynomials. (No, that's not the only way, but it's the simplest, and it certainly applies here.)
            $endgroup$
            – David C. Ullrich
            Jan 12 at 17:03










          • $begingroup$
            Ok so $f(A) = f(PDP^{-1}) = sum_{n=0}^infty c_n (PDP^{-1})^n=Psum_{n=0}^infty c_n (D)^n P^{-1}$?
            $endgroup$
            – daljit97
            Jan 12 at 17:27












          • $begingroup$
            @daljit97 $dots=Pf(D)P^{-1}$. Right.
            $endgroup$
            – David C. Ullrich
            Jan 12 at 17:32






          • 1




            $begingroup$
            @Yanko You need to read about the "functional calculus" somewhere - it's a long story. There's the "continuous functional calculus": If $D$ is diagonal with entriies $lambda_1,dots,lambda_n$ then $f(D)$ is diagonal with entries $f(lambda_1),dots,f(lambda_n)$. Or the holomorphic functional calculus: If $x$ is an element of any Banach algebra, $V$ is open, $sigma(x)subset V$ and $fin H(V)$ then you can define $f(x)$ using the Cauchy Integral Formula $f(x)=frac1{2pi i}int_Gamma f(z)(x-ze)^{-1},dz$ for a suitable cycle $Gamma$ (see Rudin Functional Analysis).
            $endgroup$
            – David C. Ullrich
            Jan 12 at 17:51


















          3












          $begingroup$

          If $f(A)$ is defined for every $A$ then $f$ must be an entire function: $$f(z)=sum_{n=0}^infty c_n z^n.$$Note that $$(PDP^{-1})^n=PD^nP^{-1}.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            (+1) Please let me note that the term $f(D)$ or $cos(D)$ is not even defined. You have to first express $f$ or $cos$ by a limit of polynomials in order to give it a meaning.
            $endgroup$
            – Yanko
            Jan 12 at 17:00








          • 1




            $begingroup$
            @Yanko Huh??? A power series is a limit of polynomials. (No, that's not the only way, but it's the simplest, and it certainly applies here.)
            $endgroup$
            – David C. Ullrich
            Jan 12 at 17:03










          • $begingroup$
            Ok so $f(A) = f(PDP^{-1}) = sum_{n=0}^infty c_n (PDP^{-1})^n=Psum_{n=0}^infty c_n (D)^n P^{-1}$?
            $endgroup$
            – daljit97
            Jan 12 at 17:27












          • $begingroup$
            @daljit97 $dots=Pf(D)P^{-1}$. Right.
            $endgroup$
            – David C. Ullrich
            Jan 12 at 17:32






          • 1




            $begingroup$
            @Yanko You need to read about the "functional calculus" somewhere - it's a long story. There's the "continuous functional calculus": If $D$ is diagonal with entriies $lambda_1,dots,lambda_n$ then $f(D)$ is diagonal with entries $f(lambda_1),dots,f(lambda_n)$. Or the holomorphic functional calculus: If $x$ is an element of any Banach algebra, $V$ is open, $sigma(x)subset V$ and $fin H(V)$ then you can define $f(x)$ using the Cauchy Integral Formula $f(x)=frac1{2pi i}int_Gamma f(z)(x-ze)^{-1},dz$ for a suitable cycle $Gamma$ (see Rudin Functional Analysis).
            $endgroup$
            – David C. Ullrich
            Jan 12 at 17:51
















          3












          3








          3





          $begingroup$

          If $f(A)$ is defined for every $A$ then $f$ must be an entire function: $$f(z)=sum_{n=0}^infty c_n z^n.$$Note that $$(PDP^{-1})^n=PD^nP^{-1}.$$






          share|cite|improve this answer









          $endgroup$



          If $f(A)$ is defined for every $A$ then $f$ must be an entire function: $$f(z)=sum_{n=0}^infty c_n z^n.$$Note that $$(PDP^{-1})^n=PD^nP^{-1}.$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 12 at 16:22









          David C. UllrichDavid C. Ullrich

          61.3k43994




          61.3k43994












          • $begingroup$
            (+1) Please let me note that the term $f(D)$ or $cos(D)$ is not even defined. You have to first express $f$ or $cos$ by a limit of polynomials in order to give it a meaning.
            $endgroup$
            – Yanko
            Jan 12 at 17:00








          • 1




            $begingroup$
            @Yanko Huh??? A power series is a limit of polynomials. (No, that's not the only way, but it's the simplest, and it certainly applies here.)
            $endgroup$
            – David C. Ullrich
            Jan 12 at 17:03










          • $begingroup$
            Ok so $f(A) = f(PDP^{-1}) = sum_{n=0}^infty c_n (PDP^{-1})^n=Psum_{n=0}^infty c_n (D)^n P^{-1}$?
            $endgroup$
            – daljit97
            Jan 12 at 17:27












          • $begingroup$
            @daljit97 $dots=Pf(D)P^{-1}$. Right.
            $endgroup$
            – David C. Ullrich
            Jan 12 at 17:32






          • 1




            $begingroup$
            @Yanko You need to read about the "functional calculus" somewhere - it's a long story. There's the "continuous functional calculus": If $D$ is diagonal with entriies $lambda_1,dots,lambda_n$ then $f(D)$ is diagonal with entries $f(lambda_1),dots,f(lambda_n)$. Or the holomorphic functional calculus: If $x$ is an element of any Banach algebra, $V$ is open, $sigma(x)subset V$ and $fin H(V)$ then you can define $f(x)$ using the Cauchy Integral Formula $f(x)=frac1{2pi i}int_Gamma f(z)(x-ze)^{-1},dz$ for a suitable cycle $Gamma$ (see Rudin Functional Analysis).
            $endgroup$
            – David C. Ullrich
            Jan 12 at 17:51




















          • $begingroup$
            (+1) Please let me note that the term $f(D)$ or $cos(D)$ is not even defined. You have to first express $f$ or $cos$ by a limit of polynomials in order to give it a meaning.
            $endgroup$
            – Yanko
            Jan 12 at 17:00








          • 1




            $begingroup$
            @Yanko Huh??? A power series is a limit of polynomials. (No, that's not the only way, but it's the simplest, and it certainly applies here.)
            $endgroup$
            – David C. Ullrich
            Jan 12 at 17:03










          • $begingroup$
            Ok so $f(A) = f(PDP^{-1}) = sum_{n=0}^infty c_n (PDP^{-1})^n=Psum_{n=0}^infty c_n (D)^n P^{-1}$?
            $endgroup$
            – daljit97
            Jan 12 at 17:27












          • $begingroup$
            @daljit97 $dots=Pf(D)P^{-1}$. Right.
            $endgroup$
            – David C. Ullrich
            Jan 12 at 17:32






          • 1




            $begingroup$
            @Yanko You need to read about the "functional calculus" somewhere - it's a long story. There's the "continuous functional calculus": If $D$ is diagonal with entriies $lambda_1,dots,lambda_n$ then $f(D)$ is diagonal with entries $f(lambda_1),dots,f(lambda_n)$. Or the holomorphic functional calculus: If $x$ is an element of any Banach algebra, $V$ is open, $sigma(x)subset V$ and $fin H(V)$ then you can define $f(x)$ using the Cauchy Integral Formula $f(x)=frac1{2pi i}int_Gamma f(z)(x-ze)^{-1},dz$ for a suitable cycle $Gamma$ (see Rudin Functional Analysis).
            $endgroup$
            – David C. Ullrich
            Jan 12 at 17:51


















          $begingroup$
          (+1) Please let me note that the term $f(D)$ or $cos(D)$ is not even defined. You have to first express $f$ or $cos$ by a limit of polynomials in order to give it a meaning.
          $endgroup$
          – Yanko
          Jan 12 at 17:00






          $begingroup$
          (+1) Please let me note that the term $f(D)$ or $cos(D)$ is not even defined. You have to first express $f$ or $cos$ by a limit of polynomials in order to give it a meaning.
          $endgroup$
          – Yanko
          Jan 12 at 17:00






          1




          1




          $begingroup$
          @Yanko Huh??? A power series is a limit of polynomials. (No, that's not the only way, but it's the simplest, and it certainly applies here.)
          $endgroup$
          – David C. Ullrich
          Jan 12 at 17:03




          $begingroup$
          @Yanko Huh??? A power series is a limit of polynomials. (No, that's not the only way, but it's the simplest, and it certainly applies here.)
          $endgroup$
          – David C. Ullrich
          Jan 12 at 17:03












          $begingroup$
          Ok so $f(A) = f(PDP^{-1}) = sum_{n=0}^infty c_n (PDP^{-1})^n=Psum_{n=0}^infty c_n (D)^n P^{-1}$?
          $endgroup$
          – daljit97
          Jan 12 at 17:27






          $begingroup$
          Ok so $f(A) = f(PDP^{-1}) = sum_{n=0}^infty c_n (PDP^{-1})^n=Psum_{n=0}^infty c_n (D)^n P^{-1}$?
          $endgroup$
          – daljit97
          Jan 12 at 17:27














          $begingroup$
          @daljit97 $dots=Pf(D)P^{-1}$. Right.
          $endgroup$
          – David C. Ullrich
          Jan 12 at 17:32




          $begingroup$
          @daljit97 $dots=Pf(D)P^{-1}$. Right.
          $endgroup$
          – David C. Ullrich
          Jan 12 at 17:32




          1




          1




          $begingroup$
          @Yanko You need to read about the "functional calculus" somewhere - it's a long story. There's the "continuous functional calculus": If $D$ is diagonal with entriies $lambda_1,dots,lambda_n$ then $f(D)$ is diagonal with entries $f(lambda_1),dots,f(lambda_n)$. Or the holomorphic functional calculus: If $x$ is an element of any Banach algebra, $V$ is open, $sigma(x)subset V$ and $fin H(V)$ then you can define $f(x)$ using the Cauchy Integral Formula $f(x)=frac1{2pi i}int_Gamma f(z)(x-ze)^{-1},dz$ for a suitable cycle $Gamma$ (see Rudin Functional Analysis).
          $endgroup$
          – David C. Ullrich
          Jan 12 at 17:51






          $begingroup$
          @Yanko You need to read about the "functional calculus" somewhere - it's a long story. There's the "continuous functional calculus": If $D$ is diagonal with entriies $lambda_1,dots,lambda_n$ then $f(D)$ is diagonal with entries $f(lambda_1),dots,f(lambda_n)$. Or the holomorphic functional calculus: If $x$ is an element of any Banach algebra, $V$ is open, $sigma(x)subset V$ and $fin H(V)$ then you can define $f(x)$ using the Cauchy Integral Formula $f(x)=frac1{2pi i}int_Gamma f(z)(x-ze)^{-1},dz$ for a suitable cycle $Gamma$ (see Rudin Functional Analysis).
          $endgroup$
          – David C. Ullrich
          Jan 12 at 17:51




















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